What evidence do you have that this is evaluatable in any form but approximately?
–
Mariano Suárez-Alvarez♦Jul 22 '13 at 1:58

My best guess would be to use complex variable methods. Have you looked into that?
–
SuugakuJul 22 '13 at 2:04

Using trig identities, your integral reduces to determining integrals of the form $\int_0^\infty \cos(x^p)/(1+x^2) dx$ and $\int_0^\infty \sin(x^p)/(1+x^2) dx$. These seem like reasonable integrals to try to compute, perhaps someone has done them before.
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nayrbJul 22 '13 at 2:17

I started solving this insane integral. To solve the partial trigonometric integrals, I convert the common denominator into a geometric series, and the partial integrals can be solved. The results can be only be expressed in a Hypergeometric Function.
–
ArucardNov 1 '13 at 0:06

2 Answers
2

Since the integrand is bounded above and below by:
$$ \frac{16}{x^2+1} > f(x) > \frac{-14}{x^2+1}$$
Both which converge (t0 $8\pi$ and $-7\pi$ respectively), the integral converges to some value in between. Other than that, I'd bet against a closed form solution.

Not a fully answer, but hopefully useful information: for the cosine part, I noticed that $$\begin{aligned}\frac{\partial^2}{\partial x^2}I(a)&=\int_0^\infty\frac{\partial^2}{\partial x^2}\frac{\cos\left(x^\pi-a\right)}{1+x^2}\,dx\\
& = -\int_0^\infty \frac{\cos\left(x^\pi-a\right)}{1+x^2}\,dx\\
&=-I(a)\\\end{aligned}$$ and, then, our integral $I(e)$ $$I(e)=K\sin(e)$$ for some constant $K.$ One way to find $K$ is evaluating $I(\pi/2)$ or another convenient value. Numerical evidence suggests $K=-{\cos(1)+4\over \pi+1/3}$, but I don't know whether Wolfram is pointing a rational value or giving up calculations. I'd like someone to test it in a more powerful environment.